Simple Question

[dttom]

1) During the cracking reaction of heavy oil, if we place some medical paraffin in the bottom of the boiling tube and place some powdered broken porous pot in the middle of the boiling tube, what'll happen if we heat the bottom of the boiling tuble (the place where medical paraffin is placed) but not the middle (where powdered porous pot placed)?

2) How to test the flammability of a substance? And what observation can be seen?

3) What's the oxidation number of glucose,C6H12O6? zero?

[RyanJ]

3) What's the oxidation number of glucose' date='C6H12O6? zero?[/quote']

 

You can't say the oxidation number of a compound, you can say it for a "group" or a single element in that compound. If you did it would loose its meaning as the oxidation number for any compound is always zero.

 

Cheers,

 

Ryan Jones

[YT2095]

unless he means the OB on decomposition/combustion? then it would be like -12.

 

I`m also uncertain about these "porous pot" things, is for anti-bump, or is the hope to crack the parafin (already cracked) into smaller molecules by a medium such as Zeolite (also ressembling a clay material)????

[dttom]

I think I have made a silly mistake, actually I want to know the oxidation number of carbon in glucose.

And also, what about the test for flammability?

[RyanJ]

I think I have made a silly mistake' date=' actually I want to know the oxidation number of carbon in glucose.

And also, what about the test for flammability?[/quote']

 

I hate doing this for such complec molecules but none the less....

 

[ce]C6H12O6[/ce]

 

Im guessing that H has an oxidation number of -1 here otherwise Carbon has an oxidation number of 0 and we don't want that so....

 

(12 * -1) + (6 * -2) = -24

(24 / 6) = 4

 

So, Carbon has an oxidation number of 4.

 

As for the flammability test, there could be many variations on it, initial flammability, etc.

 

I am wrong, Look at Woelen's post below - I've had a bad day today

 

Cheers,

 

Ryan Jones

[woelen]

Glucose: CH2(OH)CH(OH)...CH(OH)C(O)H

 

Assume oxidation state of H equals +1 (not -1, because H is a stronger reductor than C), oxygen equals -2.

 

The last C is in a more oxidized state (it has an aldehyde group on it), than the other 5 C's. The first C has two H's and one -OH and hence is in a more reduced state than the other C's.

 

So, oxidation state of first C equals -1, oxidation state of middle four C's equals 0, and the oxidation state of the last C equals +1.

 

Glucose also exists in a ring-form (in reality there is an equilibrium between these two forms). I do not precisely recall the ring form, but most likely in that case, the oxidation state of the two end-C's will be different, but in order to determine, one has to checkup the ring structure.